Recently I am reading the J.P.May's "A Concise Course in Algebraic Topology" and here is the part that I have confusion with:
Theorem. Let $p : E \to B$ be a covering and let $f : X \to B$ be a continuous map. Choose $x \in X$, let $b = f (x)$, and choose $e \in F_{b}$ . There exists a map $g : X \to E$ such that $g(x) = e$ and $p \circ g = f$ if and only if $f_{*} (\pi_{1} (X, x)) \subset p_{*} (\pi_{1} (E, e))$ in $\pi_{1} (B, b)$. When this condition holds, there is a unique such map $g$.
Proof: If $g$ exists, its properties directly imply that $im(f_{*}) \subset im(p_{*})$. Thus assume that $im(f_{*}) \subset im(p_{*})$. Applied to the covering $\Pi(p) : \Pi(E) \to \Pi(B)$, the analogue for groupoids gives a functor $\Pi(X) → \Pi(E)$ that restricts on objects to the unique map $g : X \to E$ of sets such that $g(x) = e$ and $p \circ g = f$. We need only check that $g$ is continuous, and this holds because $p$ is a local homeomorphism. In detail, if $y \in X$ and $g(y) \in U$ , where $U$ is an open subset of $E$, then there is a smaller open neighborhood $U$ of $g(y)$ that $p$ maps homeomorphically onto an open subset $V$ of $B$. If $W$ is any path connected neighborhood of $y$ such that $f (W) \subset V$ , then $g(W) \subset U$ by inspection of the definition of $g$.
I mainly have confusion with the last sentence in the proof. How do I get $g(W) \subset U$ by inspection of the definition of $g$?
By definition, $g$ is the object function of a functor $G:\Pi(X)\to \Pi(E)$ that satisfies $\Pi(p)\circ G=\Pi(f)$. So now let $z\in W$ and let $\gamma$ be a path from $y$ to $z$ contained in $W$. Since $\Pi(p)\circ G=\Pi(f)$, $G(\gamma)$ must be the unique path in $E$ starting at $g(y)$ which is a lift of $\gamma$ along $p$. But since $p$ restricts to a homeomorphism $p|_U:U\to V$, such a lift can be constructed as just $p|_U^{-1}\circ f\circ \gamma$, and so by uniqueness $G(\gamma)=p|_U^{-1}\circ f\circ \gamma$. In particular, this means that the image of $G(\gamma)$ is contained entirely in $U$ and so its endpoint $g(z)=G(z)$ is in $U$.